3.55.94 \(\int (-18+4 x^2-4 \log (\frac {10}{x})+(-8+6 x^2-2 \log (\frac {10}{x})) \log (x^2)) \, dx\)

Optimal. Leaf size=25 \[ 20+2 \left (x+x \left (-5+x^2-\log \left (\frac {10}{x}\right )\right ) \log \left (x^2\right )\right ) \]

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Rubi [A]  time = 0.10, antiderivative size = 33, normalized size of antiderivative = 1.32, number of steps used = 11, number of rules used = 6, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.171, Rules used = {2295, 6741, 12, 6742, 2304, 2361} \begin {gather*} -2 x \log \left (\frac {10}{x}\right ) \log \left (x^2\right )-10 x \log \left (x^2\right )+2 x^3 \log \left (x^2\right )+2 x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[-18 + 4*x^2 - 4*Log[10/x] + (-8 + 6*x^2 - 2*Log[10/x])*Log[x^2],x]

[Out]

2*x - 10*x*Log[x^2] + 2*x^3*Log[x^2] - 2*x*Log[10/x]*Log[x^2]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2295

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x] /; FreeQ[{c, n}, x]

Rule 2304

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*Log[c*x^
n]))/(d*(m + 1)), x] - Simp[(b*n*(d*x)^(m + 1))/(d*(m + 1)^2), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[m, -1
]

Rule 2361

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_.) + Log[(f_.)*(x_)^(r_.)]*(e_.)), x_Symbol] :> With[{u =
IntHide[(a + b*Log[c*x^n])^p, x]}, Dist[d + e*Log[f*x^r], u, x] - Dist[e*r, Int[SimplifyIntegrand[u/x, x], x],
x]] /; FreeQ[{a, b, c, d, e, f, n, p, r}, x]

Rule 6741

Int[u_, x_Symbol] :> With[{v = NormalizeIntegrand[u, x]}, Int[v, x] /; v =!= u]

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=-18 x+\frac {4 x^3}{3}-4 \int \log \left (\frac {10}{x}\right ) \, dx+\int \left (-8+6 x^2-2 \log \left (\frac {10}{x}\right )\right ) \log \left (x^2\right ) \, dx\\ &=-22 x+\frac {4 x^3}{3}-4 x \log \left (\frac {10}{x}\right )+\int 2 \left (-4+3 x^2-\log \left (\frac {10}{x}\right )\right ) \log \left (x^2\right ) \, dx\\ &=-22 x+\frac {4 x^3}{3}-4 x \log \left (\frac {10}{x}\right )+2 \int \left (-4+3 x^2-\log \left (\frac {10}{x}\right )\right ) \log \left (x^2\right ) \, dx\\ &=-22 x+\frac {4 x^3}{3}-4 x \log \left (\frac {10}{x}\right )+2 \int \left (-4 \log \left (x^2\right )+3 x^2 \log \left (x^2\right )-\log \left (\frac {10}{x}\right ) \log \left (x^2\right )\right ) \, dx\\ &=-22 x+\frac {4 x^3}{3}-4 x \log \left (\frac {10}{x}\right )-2 \int \log \left (\frac {10}{x}\right ) \log \left (x^2\right ) \, dx+6 \int x^2 \log \left (x^2\right ) \, dx-8 \int \log \left (x^2\right ) \, dx\\ &=-6 x-4 x \log \left (\frac {10}{x}\right )-10 x \log \left (x^2\right )+2 x^3 \log \left (x^2\right )-2 x \log \left (\frac {10}{x}\right ) \log \left (x^2\right )+4 \int \left (1+\log \left (\frac {10}{x}\right )\right ) \, dx\\ &=-2 x-4 x \log \left (\frac {10}{x}\right )-10 x \log \left (x^2\right )+2 x^3 \log \left (x^2\right )-2 x \log \left (\frac {10}{x}\right ) \log \left (x^2\right )+4 \int \log \left (\frac {10}{x}\right ) \, dx\\ &=2 x-10 x \log \left (x^2\right )+2 x^3 \log \left (x^2\right )-2 x \log \left (\frac {10}{x}\right ) \log \left (x^2\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 23, normalized size = 0.92 \begin {gather*} 2 \left (x+x \left (-5+x^2-\log \left (\frac {10}{x}\right )\right ) \log \left (x^2\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[-18 + 4*x^2 - 4*Log[10/x] + (-8 + 6*x^2 - 2*Log[10/x])*Log[x^2],x]

[Out]

2*(x + x*(-5 + x^2 - Log[10/x])*Log[x^2])

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fricas [A]  time = 0.67, size = 45, normalized size = 1.80 \begin {gather*} 4 \, x \log \left (\frac {10}{x}\right )^{2} + 4 \, {\left (x^{3} - 5 \, x\right )} \log \left (10\right ) - 4 \, {\left (x^{3} + x \log \left (10\right ) - 5 \, x\right )} \log \left (\frac {10}{x}\right ) + 2 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*log(10/x)+6*x^2-8)*log(x^2)-4*log(10/x)+4*x^2-18,x, algorithm="fricas")

[Out]

4*x*log(10/x)^2 + 4*(x^3 - 5*x)*log(10) - 4*(x^3 + x*log(10) - 5*x)*log(10/x) + 2*x

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giac [A]  time = 0.18, size = 42, normalized size = 1.68 \begin {gather*} 4 \, x \log \relax (x)^{2} + 4 \, x {\left (\log \left (10\right ) + 6\right )} + 4 \, {\left (x^{3} - x {\left (\log \left (10\right ) + 6\right )}\right )} \log \relax (x) - 4 \, x \log \left (\frac {10}{x}\right ) - 22 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*log(10/x)+6*x^2-8)*log(x^2)-4*log(10/x)+4*x^2-18,x, algorithm="giac")

[Out]

4*x*log(x)^2 + 4*x*(log(10) + 6) + 4*(x^3 - x*(log(10) + 6))*log(x) - 4*x*log(10/x) - 22*x

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maple [A]  time = 0.13, size = 34, normalized size = 1.36

method	result	size
norman	\(2 x -10 x \ln \left (x^{2}\right )+2 x^{3} \ln \left (x^{2}\right )-2 x \ln \left (x^{2}\right ) \ln \left (\frac {10}{x}\right )\)	\(34\)
default	\(2 x +2 x^{3} \ln \left (x^{2}\right )-2 \ln \left (x^{2}\right ) \ln \left (10\right ) x +4 \ln \left (10\right ) x +4 x \ln \left (\frac {1}{x}\right )-10 x \ln \left (x^{2}\right )-2 \ln \left (x^{2}\right ) \ln \left (\frac {1}{x}\right ) x -4 \ln \left (\frac {10}{x}\right ) x\)	\(62\)
risch	\(4 x \ln \relax (x )^{2}+\left (-i x \pi \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )+2 i x \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2}-i x \pi \mathrm {csgn}\left (i x^{2}\right )^{3}+4 x^{3}-4 x \ln \relax (5)-4 x \ln \relax (2)-24 x \right ) \ln \relax (x )-i \pi \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right ) x^{3}-i \pi \mathrm {csgn}\left (i x^{2}\right )^{3} x^{3}+i \pi \ln \relax (5) \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right ) x -2 i \pi \ln \relax (2) \mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2} x -10 i x \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2}+5 i x \pi \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )+i \pi \ln \relax (2) \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right ) x +2 i \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2} x^{3}+i \pi \ln \relax (5) \mathrm {csgn}\left (i x^{2}\right )^{3} x +5 i x \pi \mathrm {csgn}\left (i x^{2}\right )^{3}-2 i \pi \ln \relax (5) \mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2} x +i \pi \ln \relax (2) \mathrm {csgn}\left (i x^{2}\right )^{3} x +4 x \ln \relax (5)+4 x \ln \relax (2)+2 x -4 \ln \left (\frac {10}{x}\right ) x\)	\(331\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-2*ln(10/x)+6*x^2-8)*ln(x^2)-4*ln(10/x)+4*x^2-18,x,method=_RETURNVERBOSE)

[Out]

2*x-10*x*ln(x^2)+2*x^3*ln(x^2)-2*x*ln(x^2)*ln(10/x)

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maxima [A]  time = 0.46, size = 49, normalized size = 1.96 \begin {gather*} 4 \, x {\left (\log \relax (5) + \log \relax (2) + 6\right )} + 2 \, {\left (x^{3} - x \log \left (\frac {10}{x}\right ) - 5 \, x\right )} \log \left (x^{2}\right ) - 4 \, x \log \relax (x) - 4 \, x \log \left (\frac {10}{x}\right ) - 22 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*log(10/x)+6*x^2-8)*log(x^2)-4*log(10/x)+4*x^2-18,x, algorithm="maxima")

[Out]

4*x*(log(5) + log(2) + 6) + 2*(x^3 - x*log(10/x) - 5*x)*log(x^2) - 4*x*log(x) - 4*x*log(10/x) - 22*x

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mupad [B]  time = 3.74, size = 63, normalized size = 2.52 \begin {gather*} 6\,x+4\,x\,\ln \left (\frac {1}{x}\right )-10\,x\,\ln \left (x^2\right )+4\,x\,\ln \left (10\right )+2\,x^3\,\ln \left (x^2\right )-4\,x\,\left (\ln \left (\frac {10}{x}\right )+1\right )-2\,x\,\ln \left (\frac {1}{x}\right )\,\ln \left (x^2\right )-2\,x\,\ln \left (x^2\right )\,\ln \left (10\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(4*x^2 - log(x^2)*(2*log(10/x) - 6*x^2 + 8) - 4*log(10/x) - 18,x)

[Out]

6*x + 4*x*log(1/x) - 10*x*log(x^2) + 4*x*log(10) + 2*x^3*log(x^2) - 4*x*(log(10/x) + 1) - 2*x*log(1/x)*log(x^2
) - 2*x*log(x^2)*log(10)

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sympy [A]  time = 0.25, size = 31, normalized size = 1.24 \begin {gather*} x \log {\left (x^{2} \right )}^{2} + 2 x + \left (2 x^{3} - 10 x - 2 x \log {\left (10 \right )}\right ) \log {\left (x^{2} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*ln(10/x)+6*x**2-8)*ln(x**2)-4*ln(10/x)+4*x**2-18,x)

[Out]

x*log(x**2)**2 + 2*x + (2*x**3 - 10*x - 2*x*log(10))*log(x**2)

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